# Math Help - integration

1. ## integration

Hi all
I have a problem that I've nearly finished but I'm stuck on the last bit so I'm hoping someone could help out. The question is

$\int\int (y^2 z^2 + z^2 x^2 + x^2 y^2) \: dS$
Evaluated on the region of $z^2 = x^2 + y^2$ between z=1 and z=2.

My workings so far are....
With z=√(x²+y²), the integrand should become
$y^2 (x^2 + y^2) + x^2 (x^2 + y^2) + x^2 y^2$
which simplifies to $(x^2 + y^2)^2 + x^2 y^2$

Now I have
$\sqrt{2} \int \int ( (x^2 + y^2)^2 + x^2 y^2) \: dA$

To evaluate using polar coordinates, I need to determine the limits of integration.
Since z ranges from 1 to 2 and the equation of the cone is z=r, I have $1 \leq r \leq 2$, and $0\leq \theta \leq 2\pi$

By substituting in x=r*cos(θ) and y=r*sin(θ), and multiplying by 'r' (the Jacobian determinant), I get
$\sqrt{2} \int_{0}^{2\pi} \int_{1}^{2}(r^5 + r^3 \cos(\theta)^2 \sin(\theta)^2) \: dr \: d\theta$

i think the best way is to integrate WRT r first:

$\sqrt{2} \int_{0}^{2\pi} \int_{1}^{2}(r^5 + r^3 \cos(\theta)^2 \sin(\theta)^2) \: dr \: d\theta$

(im going to leave out the $\sqrt{2}$ for now)

$\Rightarrow \int_{0}^{2\pi} [\frac{r^6}{6} + \frac{r^4}{4}]_{1}^{2} \cos(\theta)^2 \sin(\theta)^2) \: d\theta$

$\Rightarrow \int_{0}^{2\pi} [\frac{32}{3} + 4 \cos(\theta)^2 \sin(\theta)^2)] - [\frac{1}{6} + \frac{1}{4} \cos(\theta)^2 \sin(\theta)^2)] \: d\theta$

I know I then simplify this and integrate WRT theta but this is where I'm running into difficulties so any help at all would be so great.

Thanks

2. Originally Posted by michaela-donnelly
Hi all
I have a problem that I've nearly finished but I'm stuck on the last bit so I'm hoping someone could help out. The question is

$\int\int (y^2 z^2 + z^2 x^2 + x^2 y^2) \: dS$
Evaluated on the region of $z^2 = x^2 + y^2$ between z=1 and z=2.

My workings so far are....
With z=√(x²+y²), the integrand should become
$y^2 (x^2 + y^2) + x^2 (x^2 + y^2) + x^2 y^2$
which simplifies to $(x^2 + y^2)^2 + x^2 y^2$

Now I have
$\sqrt{2} \int \int ( (x^2 + y^2)^2 + x^2 y^2) \: dA$

To evaluate using polar coordinates, I need to determine the limits of integration.
Since z ranges from 1 to 2 and the equation of the cone is z=r, I have $1 \leq r \leq 2$, and $0\leq \theta \leq 2\pi$

By substituting in x=r*cos(θ) and y=r*sin(θ), and multiplying by 'r' (the Jacobian determinant), I get
$\sqrt{2} \int_{0}^{2\pi} \int_{1}^{2}(r^5 + r^3 \cos(\theta)^2 \sin(\theta)^2) \: dr \: d\theta$

i think the best way is to integrate WRT r first:

$\sqrt{2} \int_{0}^{2\pi} \int_{1}^{2}(r^5 + r^3 \cos(\theta)^2 \sin(\theta)^2) \: dr \: d\theta$

(im going to leave out the $\sqrt{2}$ for now)

$\Rightarrow \int_{0}^{2\pi} [\frac{r^6}{6} + \frac{r^4}{4}]_{1}^{2} \cos(\theta)^2 \sin(\theta)^2) \: d\theta$

$\Rightarrow \int_{0}^{2\pi} [\frac{32}{3} + 4 \cos(\theta)^2 \sin(\theta)^2)] - [\frac{1}{6} + \frac{1}{4} \cos(\theta)^2 \sin(\theta)^2)] \: d\theta$

I know I then simplify this and integrate WRT theta but this is where I'm running into difficulties so any help at all would be so great.

Thanks
could you fix the LaTeX please? we use [tex] tags here, not [tex]

3. Originally Posted by Jhevon
could you fix the LaTeX please? we use [tex] tags here, not [tex]
I knew it wasn't correct but I'd no idea, until you told me, how to change the BB Code. Thanks for the tip, much appreciated!

4. Originally Posted by michaela-donnelly
Hi all
I have a problem that I've nearly finished but I'm stuck on the last bit so I'm hoping someone could help out. The question is

$\int\int (y^2 z^2 + z^2 x^2 + x^2 y^2) \: dS$
Evaluated on the region of $z^2 = x^2 + y^2$ between z=1 and z=2.
Let $R = \{(r,\theta) \in \mathbb{R}^2 : 1\leq r\leq 2 \mbox{ and }0\leq \theta \leq 2\pi \}$. Define $\bold{G}:R\mapsto \mathbb{R}$ by $\bold{G}(r,\theta) = (r\cos\theta, r\sin \theta, r)$. Which means the integral is, $\int_0^{2\pi} \int_1^2 f(r\cos \theta, r\sin \theta, r) \left| \partial_r \bold{G} \times \partial_{\theta} \bold{G}\right| ~ dA$